SOLUTION: My problem is if the vertex of f(x)=x^2+bx+c is at (2,4), find b and c.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: My problem is if the vertex of f(x)=x^2+bx+c is at (2,4), find b and c.      Log On


   

Question 103749This question is from textbook Precalculus A GRAPHING APPROACH
: My problem is if the vertex of f(x)=x^2+bx+c is at (2,4), find b and c. This question is from textbook Precalculus A GRAPHING APPROACH

Found 2 solutions by scott8148, edjones:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
for a parabola with the vertex at (h,k) ... y=a(x-h)^2+k ... f(x)=x^2-4x+4+4 ... f(x)=x^2-4x+8

b=-4 and c=8

the coefficient of the x^2 term in the original equation is 1 so a=1

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
f%5Bx%5D=a%28x-h%29%5E2%2Bk Standard form of the quadratic equation x%3C%3E0
The vertex point is (h,k).
h=2 k=4
f[x]=a(x-2)^2+4
=x^2-4x+4+4 (let a=1)
=x^2-4x+8
So, b=-4 and c=8
Ed
graph%28500%2C500%2C-10%2C10%2C-10%2C10%2Cx%5E2-4x%2B8%29